Difference between revisions of "Exercises:Mond - Topology - 1/Question 7"

Revision as of 22:08, 11 October 2016

Section B

Question 7

Let [ilmath]D^2[/ilmath] denote the closed unit disk in [ilmath]\mathbb{R}^2[/ilmath] and define an equivalence relation on [ilmath]D^2[/ilmath] by setting [ilmath]x_1\sim x_2[/ilmath] if [ilmath]\Vert x_1\Vert=\Vert x_2\Vert=1[/ilmath] ("collapsing the boundary to a single point"). Show that [ilmath]\frac{D^2}{\sim} [/ilmath] is homeomorphic to [ilmath]\mathbb{S}^2[/ilmath] - the sphere.

• Hint: first define a surjection [ilmath](:D^2\rightarrow\mathbb{S}^2)[/ilmath] mapping all of [ilmath]\partial D^2[/ilmath] to the north pole. This may be defined using a good picture or a formula.

Solution

The idea is to double the radius of [ilmath]D^2[/ilmath], then pop it out into a hemisphere, then pull the rim to a point

Picture showing the "expanding [ilmath]D^2[/ilmath]", the embedding-in-[ilmath]\mathbb{R}^3[/ilmath] part, and the "popping out"

Definitions:

• [ilmath]H[/ilmath] denotes the hemisphere in my picture.
• [ilmath]E:D^2\rightarrow H[/ilmath] is the composition of maps in my diagram that take [ilmath]D^2[/ilmath], double its radius, then embed it in [ilmath]\mathbb{R}^3[/ilmath] then "pop it out" into a hemisphere. We take it as obvious that it is a homeomorphism
• [ilmath]f':H\rightarrow\mathbb{S}^2[/ilmath], this is the map in the top picture. It takes the hemisphere and pulls the boundary/rim in (along the blue lines) to the north pole of the red sphere. [ilmath]f'(\partial H)=(0,0,1)\in\mathbb{R}^3[/ilmath], it should be clear that for all [ilmath]x\in H-\partial H[/ilmath] that [ilmath]f'(x)[/ilmath] is intended to be a point on the red sphere and that [ilmath]f'\big\vert_{H-\partial H}[/ilmath] is injective. It is also taken as clear that [ilmath]f'[/ilmath] is surjective
• Note: Click the pictures for a larger version
• [ilmath]\frac{D^2}{\sim} [/ilmath] and [ilmath]D^2/\sim[/ilmath] denote the quotient space, with this definition we get a canonical projection, [ilmath]\pi:D^2\rightarrow D^2/\sim[/ilmath] given by [ilmath]\pi:x\mapsto [x][/ilmath] where [ilmath][x][/ilmath] denotes the equivalence class of [ilmath]x[/ilmath]
• Lastly, we define [ilmath]f:D^2\rightarrow\mathbb{S}^2[/ilmath] to be the composition of [ilmath]E[/ilmath] and [ilmath]f'[/ilmath], that is: [ilmath]f:=f'\circ E[/ilmath], meaning [ilmath]f:x\mapsto f'(E(x))[/ilmath]

The situation is shown diagramatically below:

• [ilmath]\xymatrix{ D^2 \ar[d]_\pi \ar[r]^E \ar@/^1.5pc/[rr]^{f:=f'\circ E} & H \ar[r]^{f'} & \mathbb{S}^2 \\ \frac{D^2}{\sim} }[/ilmath]

Outline of the solution:

• We then want apply the passing to the quotient theorem to yield a commutative diagram: [ilmath]\xymatrix{ D^2 \ar[d]_\pi \ar[r]^E \ar@/^1.5pc/[rr]^{f} & H \ar[r]^{f'} & \mathbb{S}^2 \\ \frac{D^2}{\sim} \ar@{.>}[urr]_{\bar{f} } }[/ilmath]
  • The commutative diagram part merely means that [ilmath]f=\bar{f}\circ\pi[/ilmath]^{[Note 1]}. We get [ilmath]f=\bar{f}\circ\pi[/ilmath] as a result of the passing-to-the-quotient theorem.
  • We take this diagram as showing morphisms in the TOP category, meaning all arrows shown represent continuous maps. (Obviously...)
• Lastly, we will show that [ilmath]\bar{f} [/ilmath] is a homeomorphism using the compact-to-Hausdorff theorem

Solution body

First we must show the requirements for applying passing to the quotient are satisfied.

• We know already the maps involved are continuous and that [ilmath]\pi[/ilmath] is a quotient map. We only need to show:
  • [ilmath]f[/ilmath] is constant on the fibres of [ilmath]\pi[/ilmath], which is equivalent to:
    • [ilmath]\forall x,y\in D^2[\pi(x)=\pi(y)\implies f(x)=f(y)][/ilmath]
• Let us show this remaining condition:
  • Let [ilmath]x,y\in D^2[/ilmath] be given.
    • Suppose [ilmath]\pi(x)\ne\pi(y)[/ilmath], then by the nature of logical implication the implication is true regardless of [ilmath]f(x)[/ilmath] and [ilmath]f(y)[/ilmath]'s equality. We're done in this case.
    • Suppose [ilmath]\pi(x)=\pi(y)[/ilmath], we must show that in this case [ilmath]f(x)=y(y)[/ilmath].
      • Suppose [ilmath]x\in D^2-\partial D^2[/ilmath] (meaning [ilmath]x\in D^2[/ilmath] but [ilmath]x\notin \partial D^2[/ilmath], ie [ilmath]-[/ilmath] denotes relative complement)
        • In this case we must have [ilmath]x=y[/ilmath], as otherwise we'd not have [ilmath]\pi(x)=\pi(y)[/ilmath] (for [ilmath]x\in D^2-\partial D^2[/ilmath] we have [ilmath]\pi(x)=[x]=\{x\}[/ilmath], that is that the equivalence classes are singletons. So if [ilmath]\pi(x)=\pi(y)[/ilmath] we must have [ilmath]\pi(y)=[y]=\{x\}=[x]=\pi(x)[/ilmath]; so [ilmath]y[/ilmath] can only be [ilmath]x[/ilmath])
        • If [ilmath]x=y[/ilmath] then by the nature of [ilmath]f[/ilmath] being a function we must have [ilmath]f(x)=f(y)[/ilmath], we're done in this case
      • Suppose [ilmath]x\in \partial D^2[/ilmath] (the only case not covered) and [ilmath]\pi(x)=\pi(y)[/ilmath], we must show [ilmath]f(x)=f(y)[/ilmath]
        • Clearly if [ilmath]x\in\partial D^2[/ilmath] and [ilmath]\pi(x)=\pi(y)[/ilmath] we must have [ilmath]y\in\partial D^2[/ilmath].
          • [ilmath]E(x)[/ilmath] is mapped to the boundary/rim of [ilmath]H[/ilmath], as is [ilmath]E(y)[/ilmath] and [ilmath]f'(\text{any point on the rim of }H)=(0,0,1)\in\mathbb{R}^3[/ilmath]
          • Thus [ilmath]f'(E(x))=f'(E(y))[/ilmath], but [ilmath]f'(E(x))[/ilmath] is the very definition of [ilmath]f(x)[/ilmath], so clearly:
            • [ilmath]f(x)=f(y)[/ilmath] as required.

We may now apply the passing to the quotient theorem. This yields:

• A continuous map, [ilmath]\bar{f}:D^2/\sim\rightarrow\mathbb{S}^2[/ilmath] where [ilmath]f=\bar{f}\circ\pi[/ilmath]

In order to apply the compact-to-Hausdorff theorem and show [ilmath]\bar{f} [/ilmath] is a homeomorphism we must show it is continuous and bijective. We already have continuity (as a result of the passing-to-the-quotient theorem), we must show it is bijective.

Notes

1. ↑ Technically a diagram is said to commute if all paths through it yield equal compositions, this means that we also require [ilmath]f=f'\circ E[/ilmath], which we already have by definition of [ilmath]f[/ilmath]!

References